Solution
= Solution
On $0<x<\pi$, $\sqrt{\log y}=\sqrt{\sin^2x}=\sin x>0$. The <chain rule> gives
$$
\frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}
=\frac{1}{2y\sin x}\,y\sin2x.
$$
Using $\sin2x=2\sin x\cos x$ yields
$$
\boxed{\frac{dz}{dx}=\cos x}.
$$
= Solution
On $0<x<\pi$, $\sqrt{\log y}=\sqrt{\sin^2x}=\sin x>0$. The <chain rule> gives
$$
\frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}
=\frac{1}{2y\sin x}\,y\sin2x.
$$
Using $\sin2x=2\sin x\cos x$ yields
$$
\boxed{\frac{dz}{dx}=\cos x}.
$$